# Prove that cot A + cot (60 + A) –  cot (60 – A) = 3 cot 3A.

## Solution :

We have,

L.H.S = cot A + cot (60 + A) –  cot (60 – A)

$$\implies$$ L.H.S = $$1\over tan A$$ + $$1\over tan (60 + A)$$ – $$1\over tan (60 – A)$$

$$\implies$$ L.H.S = $$1\over tan A$$+ $$1 – \sqrt{3} tan A\over \sqrt{3} + tan A$$ – $$1 + \sqrt{3} tan A\over \sqrt{3} – tan A$$

[ By using this formula, tan (A + B) = $$tan A + tan B\over 1 – tan A tan B$$  above ]

$$\implies$$ L.H.S = $$1\over tan A$$ – $$8 tan A\over 3 – tan^2 A$$

$$\implies$$ L.H.S = $$3 – 9 tan^2 A\over 3 tan A – tan^3 A$$

L.H.S = 3($$1 – 3 tan^2 A\over 3 tan A – tan^3 A$$)

Since $$3 tan A – tan^3 A\over 1 – 3 tan^2 A$$ = tan 3A

$$\implies$$ L.H.S = $$3\over tan 3A$$ = 3 cot 3A = R.H.S